The displacement of a particle along the x axis is given by x=a ^2...

For a particle executing SHM acceleration (a) $\propto-\omega^2$ displacement $(x)$ Given $\mathrm{x}=\mathrm{a} \sin ^2 \omega \mathrm{t}$ Differentiating the above equation w.r.t, we get $\frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{a} \omega(\sin \omega \mathrm{t})(\cos \omega \mathrm{t})$ Again differentiating, we get $$\begin{aligned}& \frac{\mathrm{d}^2 \mathrm{x}}{\mathrm{dt}^2}=\mathrm{a}=2 \mathrm{a} \omega^2\left[\cos ^2 \omega \mathrm{t}-\sin ^2 \omega \mathrm{t}\right]\\ & =2 \mathrm{a} \omega^2 \cos 2 \omega \mathrm{t} \end{aligned}$$The given equation does not satisfy the condition for SHM [Eq. (i)] . Therefore, motion is not simple harmonic.